You're planning a PoE deployment. You picked the switch, you picked the cameras, and you're about to order a thousand feet of cable. But which cable — Cat5e or Cat6? Does the gauge actually matter that much for power delivery?
Yes. It matters more than most people think.
A 90-meter run of Cat5e (24 AWG) delivering 51W to a PoE++ device loses significantly more voltage than the same run on Cat6 (23 AWG). Enough to push you below the minimum operating voltage on a hot day? Maybe. Let's find out.
This article walks through the math behind PoE voltage drop, shows how cable gauge and distance interact, and gives you practical examples so you can make informed decisions about your cable plant.
The Fundamentals
PoE delivers DC power over copper conductors in Ethernet cable. Those conductors have resistance, and resistance causes voltage drop according to Ohm's law:
Vdrop = I × R
Where:
Vdrop= voltage lost in the cable (volts)I= current flowing through the cable (amps)R= total loop resistance of the cable (ohms)
The power lost in the cable is:
Ploss = I² × R
Both equations are simple. The complexity comes from figuring out the correct resistance and current values for your specific PoE setup.
Cable Resistance by AWG Gauge
The DC resistance of copper wire depends on the cross-sectional area, which is determined by the AWG (American Wire Gauge) number. Smaller AWG = thicker wire = less resistance.
| AWG | Diameter (mm) | Resistance (Ω/km) | Resistance (Ω/100m) |
|---|---|---|---|
| 26 | 0.405 | 134.5 | 13.45 |
| 24 | 0.511 | 84.2 | 8.42 |
| 23 | 0.574 | 66.8 | 6.68 |
| 22 | 0.643 | 53.0 | 5.30 |
| 20 | 0.813 | 33.3 | 3.33 |
These are per-conductor values at 20°C. Standard Cat5e uses 24 AWG. Standard Cat6 and Cat6A use 23 AWG.
Temperature Correction
Cable resistance increases with temperature. If your cables run through hot spaces (attics, industrial plants, direct sun exposure), you need to account for this.
Copper's temperature coefficient: α = 0.00393 /°C
R(T) = R(20°C) × [1 + α × (T - 20)]
| Temperature | Resistance Multiplier |
|---|---|
| 20°C | 1.00× |
| 40°C | 1.08× |
| 60°C | 1.16× |
| 75°C | 1.22× |
A cable run that measures 8Ω at 20°C becomes 9.3Ω at 60°C. That's a 16% increase in resistance — and a 16% increase in voltage drop.
How PoE Cable Resistance Works
PoE delivers power using the data pairs (or spare pairs) in the Ethernet cable. The current path depends on the PoE type:
2-Pair PoE (802.3af, 802.3at)
Power flows over two pairs (4 conductors). Current goes out on two conductors and returns on the other two. The effective loop resistance per meter is:
Rloop = 2 × Rconductor / 2 = Rconductor
(The factor of 2 is for the round trip, divided by 2 because two wires share the current in each direction.)
So for Cat5e (24 AWG): Rloop = 0.0842 Ω/m
4-Pair PoE (802.3bt Type 3 and Type 4)
All four pairs carry power, further reducing effective resistance:
Rloop = 2 × Rconductor / 4 = Rconductor / 2
For Cat5e (24 AWG) with 4-pair: Rloop = 0.0421 Ω/m
This is why PoE++ (802.3bt) uses all four pairs — you need the lower resistance to deliver higher power without excessive voltage drop.
Calculation Examples
Example 1: PoE Camera on Cat5e at 75 Meters
Setup:
- Standard: IEEE 802.3at (PoE+)
- PSE output: 50V DC
- PD power requirement: 12W (typical IP camera)
- Cable: Cat5e (24 AWG), 75 meters
- Temperature: 20°C
Step 1: Calculate current
I = P / V = 12W / 50V = 0.24A
Step 2: Calculate loop resistance
Rloop = 0.0842 Ω/m × 75m = 6.32Ω
Step 3: Calculate voltage drop
Vdrop = 0.24A × 6.32Ω = 1.52V
Step 4: Check PD voltage
V_PD = 50V - 1.52V = 48.48V
Minimum PD voltage for 802.3at is 42.5V. We're well above that. No problem.
Power loss in cable:
Ploss = 0.24² × 6.32 = 0.36W
That's only 3% of the delivered power. Very efficient.
Example 2: PoE++ Access Point on Cat5e at 100 Meters
Setup:
- Standard: IEEE 802.3bt Type 4
- PSE output: 52V DC
- PD power requirement: 60W (WiFi 6E access point)
- Cable: Cat5e (24 AWG), 100 meters
- Temperature: 45°C (hot ceiling space)
Step 1: Temperature-adjusted resistance
R(45°C) = 0.0842 × [1 + 0.00393 × 25] = 0.0842 × 1.098 = 0.0925 Ω/m
Step 2: Loop resistance (4-pair)
Rloop = 0.0925 / 2 × 100 = 4.63Ω
Step 3: Current
I = 60 / 52 = 1.15A
Step 4: Voltage drop
Vdrop = 1.15 × 4.63 = 5.33V
Step 5: PD voltage
V_PD = 52 - 5.33 = 46.67V
Minimum for 802.3bt Type 4 is 41.1V. Margin is 5.57V. Works, but tighter than you'd like.
Power loss:
Ploss = 1.15² × 4.63 = 6.12W
That's 10.2% of the delivered power wasted as heat in the cable. Not great, but within spec.
Example 3: Same Setup on Cat6 (23 AWG)
With Cat6 at 45°C:
R(45°C) = 0.0668 × 1.098 = 0.0733 Ω/m
Rloop = 0.0733 / 2 × 100 = 3.67Ω
Vdrop = 1.15 × 3.67 = 4.22V
V_PD = 52 - 4.22 = 47.78V
Ploss = 1.15² × 3.67 = 4.85W
| Metric | Cat5e (24 AWG) | Cat6 (23 AWG) | Improvement |
|---|---|---|---|
| Voltage drop | 5.33V | 4.22V | -21% |
| PD voltage | 46.67V | 47.78V | +1.11V |
| Power loss | 6.12W | 4.85W | -21% |
| Margin above minimum | 5.57V | 6.68V | +20% |
The thicker gauge gives you an extra volt of margin and saves over 1W of power loss. At 100 meters with high-power devices, that's meaningful.
When Voltage Drop Becomes a Real Problem
Voltage drop starts causing issues when:
- The PD reboots under load — when the access point transmits at full power, current spikes, voltage drops further, and the PD browns out.
- IR camera heaters fail — thermal cameras with built-in heaters can draw 20W+ extra in cold weather, pushing the power budget past what the cable can deliver.
- PTZ cameras stall during movement — motor inrush current creates momentary voltage sag.
- The PSE reports faults — managed switches track per-port power and will shut down ports that exceed their budget.
If you're seeing any of these symptoms on long cable runs, voltage drop is the prime suspect.
Cable Selection Guide by Use Case
| Use Case | Power | Distance | Recommended Cable |
|---|---|---|---|
| IP camera (indoor) | <13W | <80m | Cat5e is fine |
| IP camera (outdoor w/ heater) | 25-40W | <100m | Cat6 recommended |
| WiFi access point | 25-60W | <80m | Cat6 minimum |
| WiFi 6E / high-power AP | 50-71W | <60m | Cat6A with 23AWG |
| PoE lighting | 30-60W | <100m | Cat6 minimum |
| Industrial sensor | <13W | <100m | Cat5e is fine |
| PTZ camera | 30-50W | <80m | Cat6 recommended |
These are conservative guidelines. Run the actual numbers for your specific deployment.
Don't Forget the Patch Cables
Your 90-meter horizontal run might be Cat6, but if there are two 3-meter Cat5e patch cables (one at each end), the overall resistance increases. Every connector and patch cable adds resistance.
For long runs, use the same gauge throughout the entire channel, including patch cables. Mixing Cat5e patch cables into a Cat6 channel defeats the purpose of spending extra on the horizontal run.
Do the Math Before You Pull Cable
Returning to the original question — Cat5e or Cat6? — the answer depends on your power level, distance, and temperature. For low-power devices under 80 meters, Cat5e is fine. For anything pushing 50W+ at 80+ meters, especially in warm environments, Cat6 is the safer choice.
Want to run these calculations for your exact setup? Try the PoE Power Budget Planner — enter your cable gauge, distance, power level, and temperature, and get instant voltage drop and power loss numbers. Runs in your browser, exports results as PDF.
Measure twice, pull once. The math is free; re-pulling cable through conduit is not.